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uranmaximum [27]

3 years ago

11

Wo gravitational forces act on a particle, in perpendicular directions. to find the net force, can we add the magnitudes of thos

e two forces?

1 answer:

son4ous [18]3 years ago

3 0

No, you can not.

$f=mxa$

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What happens if an air conditioner is used in a house that is not well insulated and not well scaled? Explain in terms of energy

stira [4]

It will use a lot more energy (electricity) to cool down the room. Because heat energy from outside the room can easily transfer into the room again if the room is not well insulated. So more energy is needed to cool down the room again

4 0

3 years ago

Ph11_UnitPacket2019

frozen [14]

Let's see

Use snells law

$\\ \rm\Rrightarrow \dfrac{n_1}{n_2}=\dfrac{sini}{sinr}$

$\\ \rm\Rrightarrow \mu=\dfrac{sin30}{sin19.9}$

$\\ \rm\Rrightarrow \mu=0.5/0.34$

$\\ \rm\Rrightarrow \mu=1.47$

It may be glass

3 0

3 years ago

Which property describes the distance between similar points on a wave?

grin007 [14]

Answer:

B

Explanation:

5 0

3 years ago

Read 2 more answers

An electron enters a region of uniform electric field with an initial velocity of 64 km/s in the same direction as the electric

8090 [49]

Answer:

1.) 11 km/s

2.) 9.03 × 10^-5 metres

Explanation:

Given that an electron enters a region of uniform electric field with an initial velocity of 64 km/s in the same direction as the electric field, which has magnitude E = 48 N/C.

Electron q = 1.6×10^-19 C

Electron mass = 9.11×10^-31 Kg

(a) What is the speed of the electron 1.3 ns after entering this region?

E = F/q

F = Eq

Ma = Eq

M × V/t = Eq

Substitute all the parameters into the formula

9.11×10^-31 × V/1.3×10^-9 = 48 × 1.6×10^-19

V = 7.68×10^-18 /7.0×10^-22

V = 10971.43 m/s

V = 11 Km/s approximately

(b) How far does the electron travel during the 1.3 ns interval?

The initial velocity U = 64 km/s

S = ut + 1/2at^2

S = 64000×1.3×10^-6 + 1/2 × 8.4×10^12 × ( 1.3×10^-9)^2

S =8.32×10^-5 + 7.13×10^-6

S = 9.03 × 10^-5 metres

3 0

3 years ago

Tests reveal that a normal driver takes about 0.75 s before he or she can react to a situation to avoid a collision. It takes ab

nata0808 [166]

To solve this problem we will apply the linear motion kinematic equations. On these equations we will define the speed as the distance traveled in a space of time, and that speed will be in charge of indicating the reaction rate of the individual. In turn, using the ratio of speed, position and acceleration, we will clear the position and determine the distance necessary for braking.

The relation to express the velocity in terms of position for constant acceleration is as follows

$v^2 = u^2+2a(s-s_0)$

Here,

u = Initial velocity

v= Final velocity

a = Acceleration

$s_0$ = Initial position

s = Final position

PART 1) Calculate the displacement within the reaction time

$d = vt$

$d = (44)(0.75)$

$d = 33ft$

In this case we can calculate the shortest stopping distance

$0^2 = 44^2+2(-2)(s-33)$

$s = 517ft$

PART 2)

PART 1) Calculate the displacement within the reaction time

$d = vt$

$d = (44)(3)$

$d = 132ft$

In this case we can calculate the shortest stopping distance

$0^2 = 44^2+2(-2)(s-132)$

$s = 616ft$

While a person without alcohol would cost 517ft to slow down, under alcoholic substances that distance would be 616ft

7 0

4 years ago